3.108 \(\int (a+b \tanh ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=101 \[ a x+\frac{b \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac{b \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 c^{2/3} x^2+1}{\sqrt{3}}\right )}{2 \sqrt [3]{c}}+b x \tanh ^{-1}\left (c x^3\right ) \]

[Out]

a*x + (Sqrt[3]*b*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/(2*c^(1/3)) + b*x*ArcTanh[c*x^3] + (b*Log[1 - c^(2/3)*x^
2])/(2*c^(1/3)) - (b*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/(4*c^(1/3))

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Rubi [A]  time = 0.0966456, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {6091, 275, 292, 31, 634, 617, 204, 628} \[ a x+\frac{b \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac{b \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{2 c^{2/3} x^2+1}{\sqrt{3}}\right )}{2 \sqrt [3]{c}}+b x \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTanh[c*x^3],x]

[Out]

a*x + (Sqrt[3]*b*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/(2*c^(1/3)) + b*x*ArcTanh[c*x^3] + (b*Log[1 - c^(2/3)*x^
2])/(2*c^(1/3)) - (b*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/(4*c^(1/3))

Rule 6091

Int[ArcTanh[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTanh[c*x^n], x] - Dist[c*n, Int[x^n/(1 - c^2*x^(2*n)), x]
, x] /; FreeQ[{c, n}, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=a x+b \int \tanh ^{-1}\left (c x^3\right ) \, dx\\ &=a x+b x \tanh ^{-1}\left (c x^3\right )-(3 b c) \int \frac{x^3}{1-c^2 x^6} \, dx\\ &=a x+b x \tanh ^{-1}\left (c x^3\right )-\frac{1}{2} (3 b c) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x^3} \, dx,x,x^2\right )\\ &=a x+b x \tanh ^{-1}\left (c x^3\right )-\frac{1}{2} \left (b \sqrt [3]{c}\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^{2/3} x} \, dx,x,x^2\right )+\frac{1}{2} \left (b \sqrt [3]{c}\right ) \operatorname{Subst}\left (\int \frac{1-c^{2/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=a x+b x \tanh ^{-1}\left (c x^3\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac{b \operatorname{Subst}\left (\int \frac{c^{2/3}+2 c^{4/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )}{4 \sqrt [3]{c}}+\frac{1}{4} \left (3 b \sqrt [3]{c}\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=a x+b x \tanh ^{-1}\left (c x^3\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac{b \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 c^{2/3} x^2\right )}{2 \sqrt [3]{c}}\\ &=a x+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{1+2 c^{2/3} x^2}{\sqrt{3}}\right )}{2 \sqrt [3]{c}}+b x \tanh ^{-1}\left (c x^3\right )+\frac{b \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac{b \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}}\\ \end{align*}

Mathematica [A]  time = 0.042819, size = 136, normalized size = 1.35 \[ a x-\frac{b \left (\log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )+\log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )-2 \log \left (1-\sqrt [3]{c} x\right )-2 \log \left (\sqrt [3]{c} x+1\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x-1}{\sqrt{3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )\right )}{4 \sqrt [3]{c}}+b x \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTanh[c*x^3],x]

[Out]

a*x + b*x*ArcTanh[c*x^3] - (b*(-2*Sqrt[3]*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]] + 2*Sqrt[3]*ArcTan[(1 + 2*c^(1/3)
*x)/Sqrt[3]] - 2*Log[1 - c^(1/3)*x] - 2*Log[1 + c^(1/3)*x] + Log[1 - c^(1/3)*x + c^(2/3)*x^2] + Log[1 + c^(1/3
)*x + c^(2/3)*x^2]))/(4*c^(1/3))

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Maple [A]  time = 0.006, size = 99, normalized size = 1. \begin{align*} ax+bx{\it Artanh} \left ( c{x}^{3} \right ) +{\frac{b}{2\,c}\ln \left ({x}^{2}-\sqrt [3]{{c}^{-2}} \right ){\frac{1}{\sqrt [3]{{c}^{-2}}}}}-{\frac{b}{4\,c}\ln \left ({x}^{4}+\sqrt [3]{{c}^{-2}}{x}^{2}+ \left ({c}^{-2} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-2}}}}}+{\frac{b\sqrt{3}}{2\,c}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{{x}^{2}}{\sqrt [3]{{c}^{-2}}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{-2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*arctanh(c*x^3),x)

[Out]

a*x+b*x*arctanh(c*x^3)+1/2*b/c/(1/c^2)^(1/3)*ln(x^2-(1/c^2)^(1/3))-1/4*b/c/(1/c^2)^(1/3)*ln(x^4+(1/c^2)^(1/3)*
x^2+(1/c^2)^(2/3))+1/2*b*3^(1/2)/c/(1/c^2)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c^2)^(1/3)*x^2+1))

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Maxima [A]  time = 1.4484, size = 143, normalized size = 1.42 \begin{align*} \frac{1}{4} \,{\left (c{\left (\frac{2 \, \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{3}}{\left (2 \, x^{2} + \frac{1}{c^{2}}^{\frac{1}{3}}\right )}\right )}{c^{2}} - \frac{{\left (c^{2}\right )}^{\frac{1}{3}} \log \left (x^{4} + \frac{1}{c^{2}}^{\frac{1}{3}} x^{2} + \frac{1}{c^{2}}^{\frac{2}{3}}\right )}{c^{2}} + \frac{2 \,{\left (c^{2}\right )}^{\frac{1}{3}} \log \left (x^{2} - \frac{1}{c^{2}}^{\frac{1}{3}}\right )}{c^{2}}\right )} + 4 \, x \operatorname{artanh}\left (c x^{3}\right )\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arctanh(c*x^3),x, algorithm="maxima")

[Out]

1/4*(c*(2*sqrt(3)*(c^2)^(1/3)*arctan(1/3*sqrt(3)*(c^2)^(1/3)*(2*x^2 + (c^(-2))^(1/3)))/c^2 - (c^2)^(1/3)*log(x
^4 + (c^(-2))^(1/3)*x^2 + (c^(-2))^(2/3))/c^2 + 2*(c^2)^(1/3)*log(x^2 - (c^(-2))^(1/3))/c^2) + 4*x*arctanh(c*x
^3))*b + a*x

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Fricas [A]  time = 1.86935, size = 674, normalized size = 6.67 \begin{align*} \left [\frac{\sqrt{3} b c \sqrt{-\frac{1}{c^{\frac{2}{3}}}} \log \left (\frac{2 \, c^{2} x^{6} - 3 \, c^{\frac{2}{3}} x^{2} + \sqrt{3}{\left (2 \, c^{\frac{5}{3}} x^{4} - c x^{2} - c^{\frac{1}{3}}\right )} \sqrt{-\frac{1}{c^{\frac{2}{3}}}} + 1}{c^{2} x^{6} - 1}\right ) + 2 \, b c x \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a c x - b c^{\frac{2}{3}} \log \left (c^{2} x^{4} + c^{\frac{4}{3}} x^{2} + c^{\frac{2}{3}}\right ) + 2 \, b c^{\frac{2}{3}} \log \left (c x^{2} - c^{\frac{1}{3}}\right )}{4 \, c}, \frac{2 \, b c x \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, \sqrt{3} b c^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, c x^{2} + c^{\frac{1}{3}}\right )}}{3 \, c^{\frac{1}{3}}}\right ) + 4 \, a c x - b c^{\frac{2}{3}} \log \left (c^{2} x^{4} + c^{\frac{4}{3}} x^{2} + c^{\frac{2}{3}}\right ) + 2 \, b c^{\frac{2}{3}} \log \left (c x^{2} - c^{\frac{1}{3}}\right )}{4 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arctanh(c*x^3),x, algorithm="fricas")

[Out]

[1/4*(sqrt(3)*b*c*sqrt(-1/c^(2/3))*log((2*c^2*x^6 - 3*c^(2/3)*x^2 + sqrt(3)*(2*c^(5/3)*x^4 - c*x^2 - c^(1/3))*
sqrt(-1/c^(2/3)) + 1)/(c^2*x^6 - 1)) + 2*b*c*x*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a*c*x - b*c^(2/3)*log(c^2*x^4
+ c^(4/3)*x^2 + c^(2/3)) + 2*b*c^(2/3)*log(c*x^2 - c^(1/3)))/c, 1/4*(2*b*c*x*log(-(c*x^3 + 1)/(c*x^3 - 1)) +
2*sqrt(3)*b*c^(2/3)*arctan(1/3*sqrt(3)*(2*c*x^2 + c^(1/3))/c^(1/3)) + 4*a*c*x - b*c^(2/3)*log(c^2*x^4 + c^(4/3
)*x^2 + c^(2/3)) + 2*b*c^(2/3)*log(c*x^2 - c^(1/3)))/c]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*atanh(c*x**3),x)

[Out]

Timed out

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Giac [A]  time = 1.18351, size = 147, normalized size = 1.46 \begin{align*} \frac{1}{4} \,{\left (c{\left (\frac{2 \, \sqrt{3}{\left | c \right |}^{\frac{2}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + \frac{1}{{\left | c \right |}^{\frac{2}{3}}}\right )}{\left | c \right |}^{\frac{2}{3}}\right )}{c^{2}} - \frac{{\left | c \right |}^{\frac{2}{3}} \log \left (x^{4} + \frac{x^{2}}{{\left | c \right |}^{\frac{2}{3}}} + \frac{1}{{\left | c \right |}^{\frac{4}{3}}}\right )}{c^{2}} + \frac{2 \, \log \left ({\left | x^{2} - \frac{1}{{\left | c \right |}^{\frac{2}{3}}} \right |}\right )}{{\left | c \right |}^{\frac{4}{3}}}\right )} + 2 \, x \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right )\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arctanh(c*x^3),x, algorithm="giac")

[Out]

1/4*(c*(2*sqrt(3)*abs(c)^(2/3)*arctan(1/3*sqrt(3)*(2*x^2 + 1/abs(c)^(2/3))*abs(c)^(2/3))/c^2 - abs(c)^(2/3)*lo
g(x^4 + x^2/abs(c)^(2/3) + 1/abs(c)^(4/3))/c^2 + 2*log(abs(x^2 - 1/abs(c)^(2/3)))/abs(c)^(4/3)) + 2*x*log(-(c*
x^3 + 1)/(c*x^3 - 1)))*b + a*x